Let for $f(x) = 7\tan^8x + 7\tan^6x – 3\tan^4x – 3\tan^2x$, $I_1 = \int_0^{\pi/4} f(x) dx$ and $I_2 = \int_0^{\pi/4} x f(x) dx$. Then $7I_1 + 12I_2$ is equal to:
- (1) $2\pi$
- (2) $\pi$
- (3) 1
- (4) 2
Solution:
Simplifying $f(x)$:
$$ f(x) = 7\tan^6x(\tan^2x + 1) – 3\tan^2x(\tan^2x + 1) $$
$$ f(x) = (7\tan^6x – 3\tan^2x)\sec^2x $$
Calculating $I_1$:
$$ I_1 = \int_0^{\pi/4} (7\tan^6x – 3\tan^2x)\sec^2x dx $$
Let $\tan x = t \Rightarrow \sec^2x dx = dt$.
$$ I_1 = \int_0^1 (7t^6 – 3t^2) dt = [t^7 – t^3]_0^1 = (1 – 1) – 0 = 0 $$
Calculating $I_2$:
$$ I_2 = \int_0^{\pi/4} x f(x) dx = \int_0^{\pi/4} x (7\tan^6x – 3\tan^2x)\sec^2x dx $$
Using Integration by Parts ($u=x, dv=f(x)dx$). We know $\int f(x)dx = \tan^7x – \tan^3x$.
$$ I_2 = [x(\tan^7x – \tan^3x)]_0^{\pi/4} – \int_0^{\pi/4} (\tan^7x – \tan^3x) dx $$
$$ I_2 = \left[ \frac{\pi}{4}(1 – 1) – 0 \right] – \int_0^{\pi/4} \tan^3x(\tan^4x – 1) dx $$
$$ I_2 = 0 – \int_0^{\pi/4} \tan^3x(\tan^2x – 1)(\tan^2x + 1) dx $$
$$ I_2 = – \int_0^{\pi/4} (\tan^5x – \tan^3x)\sec^2x dx $$
Let $\tan x = t$:
$$ I_2 = – \int_0^1 (t^5 – t^3) dt = – \left[ \frac{t^6}{6} – \frac{t^4}{4} \right]_0^1 $$
$$ I_2 = – \left( \frac{1}{6} – \frac{1}{4} \right) = – \left( \frac{2-3}{12} \right) = \frac{1}{12} $$
Value required:
$$ 7I_1 + 12I_2 = 7(0) + 12\left(\frac{1}{12}\right) = 1 $$
Ans. (3)